Pedagogical Machine: Solving Homework Equations

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SUMMARY

The discussion focuses on solving a physics problem involving three masses (m1, m2, m3) and the force (F) applied to the system, described by the equation F = ma. The analysis concludes that for m2 to remain at rest relative to m1, the entire apparatus must share a common acceleration, leading to the derived relationship F = (m1 + m2 + m3)(m3/m2)g. The solution is validated through limiting cases, confirming that when m3 = 0, F = 0, and when m2 = 0, F approaches infinity, illustrating the physical implications of the scenario.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the concept of apparent forces in non-inertial frames
  • Knowledge of basic mechanics involving multiple masses
  • Ability to analyze limiting cases in physics problems
NEXT STEPS
  • Study the implications of non-inertial reference frames in classical mechanics
  • Explore the concept of pseudo forces and their applications in engineering
  • Investigate the dynamics of systems with multiple interacting masses
  • Learn about the mathematical derivation of forces in complex mechanical systems
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Students of physics, educators teaching mechanics, and engineers involved in dynamics and system analysis will benefit from this discussion.

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Homework Statement


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Homework Equations


F = ma


The Attempt at a Solution


What we want is for m2 to be at rest relative to m1 because this implies m2 will not be sliding across the surface of m2 and this immediately implies m3 must be at rest in the vertical direction (in the lab frame and frame of m1 because there will be no pseudo forces in the vertical direction on m3) because if m3 was not at rest vertically then m2 would be slipping along m1 while m1 was moving which is a contradiction if we assume m2 is at rest with respect to m1 while m1 is moving. Boosting to a frame co - moving with m1 we have that in this frame, m_{2}a_{2} = F_{apparent} = T - m_{2}a = 0 where a = \frac{F}{m_{1} + m_{2} + m_{3}} is the acceleration of the entire apparatus (and consequently m2 since they all move together under F). As noted above, this immediately implies m_{3}a_{3} = T - m_{3}g = 0 so combining this together we have that m_{3}g = m_{2}a so F = (m_{1} + m_{2} + m_{3})(\frac{m3}{m2})g. If I take m1 = m2 = m3 = m then F = 3mg as stated in the ans. clue. I also checked the limiting cases m3 = 0 which gives F = 0 as it should because if there is no m3 then m2 will be at rest even in the lab frame so we don't need any force on m1. If m2 = 0 then F = infinity which makes sense since if m2 = 0, m3 will go into free fall and the only way to stop it from going into free fall using a horizontal force would be an infinite one that accelerates it in the horizontal direction so fast that it doesn't get a chance to fall (this is unphysical of course). Is the solution right? Thanks!
 
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I analyzed it from the stationary frame where all masses share a common acceleration and got the same results.
I dislike working with pseudo forces because it encourages engineers to believe centrifugal forces are real.
 

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